Find $\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{3\cos^2(x)}{2-2\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $3$ (Choice B) B $\dfrac{3}{2}$ (Choice C) C $\dfrac{3}{4}$ (Choice D) D The limit doesn't exist
Answer: Substituting $x=\dfrac{\pi}{2}$ into $\dfrac{3\cos^2(x)}{2-2\sin(x)}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since our expression includes trigonometric functions, let's try to re-write it using factorization and trigonometric identities. Since we have $\cos^2(x)$ in our expression, let's rewrite it using the Pythagorean identity, $\sin^2(x)+\cos^2(x)=1$ : $\begin{aligned} &\phantom{=}\dfrac{3\cos^2(x)}{2-2\sin(x)} \\\\ &=\dfrac{3(1-\sin^2(x))}{2-2\sin(x)} \gray{\text{The Pythagorean identity}} \\\\ &=\dfrac{3(1+\sin(x))(1-\sin(x))}{2-2\sin(x)} \gray{\text{Diff. of squares}} \\\\ &=\dfrac{3(1+\sin(x))(\cancel{1-\sin(x)})}{2\cancel{(1-\sin(x))}} \gray{\text{Cancel common factors}} \\\\ &=\dfrac{3(1+\sin(x))}{2}\text{, for }x\neq \{...,- \dfrac{7\pi}{2}, - \dfrac{3\pi}{2}, \dfrac{\pi}{2}, \dfrac{5\pi}{2}, \dfrac{9\pi}{2},...\} \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $(2k+1)\dfrac{\pi}{2}$ for any integer $k$, and specifically $\dfrac{\pi}{2}$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{3\cos^2(x)}{2-2\sin(x)}=\dfrac{3(1+\sin(x))}{2}$ for all $x$ -values in the interval $(-\pi,\pi)$ except for $x=\dfrac{\pi}{2}$. Therefore, $\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{3\cos^2(x)}{2-2\sin(x)}=\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{3(1+\sin(x))}{2}=3$ (The last limit was found using direct substitution.) In conclusion, $\lim_{x\to \scriptsize\dfrac{\pi}{2}}\dfrac{3\cos^2(x)}{2-2\sin(x)}=3$.